The answer is yes. Indeed, let us write $X$ instead of $x$, according to standard notation, to distinguish between random variables (denoted by upper-case letters) and their values (denoted by lower-case letters). Let us write $P$ instead of $\Pr$, and then let us also write $A$ instead of $E$, to distinguish it from the expectation sign.
Then we need to show that \begin{equation} EX\,P(A|X)\ge P(A)\,EX \end{equation} given that \begin{equation} P(A|X>y)\ge P(A) \end{equation} for all $y$ such that $P(X>y)\ne0$.
We have \begin{multline*} EX\,P(A|X)=EX\,E(1_A|X)=EX\,1_A \\ =E\int_0^\infty dy\,1_{X>y,A}=\int_0^\infty dy\,E1_{X>y,A} \\ =\int_0^\infty dy\,P(X>y,A) =\int_0^\infty dy\,P(A|X>y)P(X>y) \\ \ge\int_0^\infty dy\,P(A)P(X>y)=P(A)\int_0^\infty dy\,P(X>y)=P(A)EX, \end{multline*} as claimed.